Answer:
Spiders cannot actually propel their bodies through the water as a swimmer does, but they can use objects to get across the water and some can run across the water.
Explanation:
Answer
given,
length of the swing = 26.2 m
inclined at an angle = 28°
let, the initial height of the Tarzan be h
h = L (1 - cos θ)
a) initial velocity v₁ = 0 m/s
final velocity of Tarzan = v_f
law of conservation of energy
PE_i + KE_i = PE_f + KE_f






= 7.75 m/s
the speed tarzan at the bottom of the swing
v_f = 7.75 m/s
b)initial speed of the = 3 m/s






v_f= 11.29 m/s
Answer:
You put the add sign, so it would be 49. I don't believe that is what you meant to put though.
Explanation: